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Question

Without expanding the determinant, prove that 

∣ ∣ ∣aa2bcbb2cacc2ab∣ ∣ ∣=∣ ∣ ∣1a2a31b2b31c2c3∣ ∣ ∣


Solution

LHS=∣ ∣ ∣aa2bcbb2cacc2ab∣ ∣ ∣
Operating R1aR1,R2bR2 and R3cR3 we obtain
=1abc∣ ∣ ∣a2a3abcb2b3abcc2c3abc∣ ∣ ∣=1abc×abc∣ ∣ ∣a2a31b2b31c2c31∣ ∣ ∣    [Taking out factor abc from C3]
=(1)2∣ ∣ ∣1a2a31b2b31c2c3∣ ∣ ∣    (using C1C3 and C2C3)
=∣ ∣ ∣1a2a31b2b31c2c3∣ ∣ ∣=RHS

Note: When we multiply any row or column of a determinant by any constant k then it is necessary that determinant divides by k.

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